使用subString在node.js中获取错误

Question

大家好，这是我在这里的第一个问题，所以我希望它没问题

我试图让我的代码获取两个之间的内容，在这个例子中我需要得到prestons

Batehaven Cl，Prestons NSW 2170，澳大利亚

我试过这段代码

var country = getCountry(data.results[0]['address_components']);
var address = data.results[3]['formatted_address'];
var suburb = address;

suburb = suburb.subString(suburb.indexOf(",")+1, suburb.lastIndexOf(","));


  console.log(address);
  console.log(suburb);

但我得到一个错误未处理拒绝TypeError：suburb.subString不是一个函数

有人可以告诉我，我犯了什么愚蠢的错误

Answer 1

Javascript方法区分大小写尝试使用substring来获取Cabramatta West一个双字的郊区示例，即下面的代码也适用于在您自己的示例中检索Prestons：

let address = "John St, Cabramatta West NSW 2166, Australia"
let suburb = address

suburb = suburb.substring(suburb.indexOf(',') + 1, suburb.lastIndexOf(','))

let states = ["TAS", "QLD", "VIC", "NSW", "SA", "WA"]
let positionOfState = -1
for (let i = 0; i < states.length; i++) {
  if (address.indexOf(states[i]) != -1) {
    positionOfState = suburb.indexOf(states[i])
    break;
  }
}
suburb = suburb.substring(1, positionOfState) // 1 is to skip the space at the start

console.log(`address -> ${address}`)
console.log(`suburb -> ${suburb}`)